CHEMISTRY COLLEGE

Water flows over Niagara Falss at the average rate of 2,400,000 kg/s, and the average height of the falls is about 50 m. Knowing that the graviatational potential energy of falling water per second = mass (kg) x height (m) x gravity (9.8 m/s2), what is the power of Niagara Falls? How many 15 W LED light bulbs could it power?

Answers

Answer 1

Answer:

1. 176 × 10^12 W ; 78400000000

Explanation:

Given the following :

Fall rate = 2,400,000kg/s

Average height of fall = 50m

Gravitational Potential of falling water = mgh = mass × acceleration due to gravity × height =

How many 15 W LED light bulbs could it power?

Recall : power = workdone / time

Workdone = gravitational potential energy

Mass of water = density * volume

Density of water = 1 * 10^3kg/m^3

Rate of fow = volume / time = 2400000

Hence,

Power = 1000 * 2,400,000 * 9.8 * 50

Power = 1176000000000

Power = 1. 176 × 10^12 W

How many 15 W LED light bulbs could it power?

1176000000000 / 15 = 78400000000

= 78400000000 15 W bulbs

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Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 50.0L tank with 14. mol of sulfur dioxide gas and 2.6 mol of oxygen gas, and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 1.6 mol. Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.

Answers

Answer:

Explanation:

From the given information:

The equation for the reaction can be represented as:

$2SO_2 + O_2 \to 2SO_3$

The I.C.E table can be represented as:

2SO₂ O₂ 2SO₃

Initial: 14 2.6 0

Change: -2x -x +2x

Equilibrium: 14 - 2x 2.6 - x 2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

then x = 1.6/2

x = 0.8 mol

For 2SO₂; we have 14 - 2x

= 14 - 2(0.8)

= 14 - 1.6

= 12.4 mol

For O₂; we have 2.6 - x

= 2.6 - 1.6

= 1.0 mol

Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

[O₂] = 1/50 = 0.02 M ,

[SO₃] = 1.6/50 = 0.032 M

Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction $2SO_2 + O_2 \to 2SO_3$ = 0.8325;

If we want to find:

$SO_2 + (1)/(2)O_2 \to SO_3$

Then:

$K_c = (0.8325)^(1/2)$

$\mathbf{K_c = 0.912}$

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

What is the potentual energy of 7 kg object, 12 m off the ground?

Answers

Answer:

823.2 J

Explanation:

PE = mgh

= (7 kg) (9.8 m/s^2) (12 m)

= 823.2 J

Answer:

RequiredAnswer:-

Mass=m=7kg

Height=h=12m

Gravitational force=g=10m/s^2

As we know that

${\boxed{\sf Potential\:energy{}_((P.E))=mgh}}$

Substitute the values

${:}\longmapsto$ $\sf P.E=7×12×10$

${:}\longmapsto$ $\sf P.E=84×10$

${:}\longmapsto$ $\sf Potential\:energy=840Joule$

. Explain why, in the sample calculations, 0.1 g of the unknown produced a GREATER freezing point depression than~e same mass of naphthalene.

Answers

Answer

Naphthalene is a non electrolyte

If the unknown compound is an electrolyte it gives 2 or more ions in solution

( NaCl >> Na+ + Cl- => 2 ions

Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)

the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )

For naphthalene

delta T = 1.86 x m

for a salt that gives 2 ions

delta T = 1.86 x m x 2

hence the lowering in freezion point of unkown is greater then napthalene

Help me plzz I need help can someone help

Answers

D volcanic corruption

A weather balloon is inflated to a volume of 27.3 L at a pressure of 738 mmHg and a temperature of 26.9 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 375 mmHg and the temperature is -15.6 ∘C.Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

Answers

Answer:

The new volume of the balloon is 46.1 L

Explanation:

Step 1: Data given

Initial volume of the balloon = 27.3 L

Initial pressure in the balloon = 738 mmHg = 0.97105 atm

Initial temperature in the balloon = 26.9 °C = 300.05 K

The pressure decreases to 375 mmHg = 0.493421 atm

The temperature lowers to -15.6 °C = 257.55 K

Step 2: Calculate the volume

P1*V1 / T1 = P2 *V2 / T2

⇒with P1 = the Initial pressure = 738 mmHg = 0.97105 atm

⇒with V1 = Initial volume of the balloon = 27.3 L

⇒with T1 = Initial temperature in the balloon = 26.9 °C = 300.05 K

⇒with P2 = the decreased pressure = 0.493421 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the lowered temperature = 257.55 K

0.97105 * 27.3 / 300.05 = 0.493421*V2 / 257.55

V2 = 46.1 L

The new volume of the balloon is 46.1 L

It is a vocal music of Mindoro

Answers

Answer:

instrumental music is used in festivals, rituals, etc.

Explanation:

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